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Dodo1 Group Title

Find the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)2=25(x^2−y^2) at the point (−3,1). The equation of this tangent line can be written in the form y=mx+b where m is:

  • one year ago
  • one year ago

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  1. Yahoo! Group Title
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    First Differentiate the function to find the slope.

    • one year ago
  2. Dodo1 Group Title
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    thank you but i dont know process. i am not sure about calculation, particulary differentiation

    • one year ago
  3. Dodo1 Group Title
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    I mean i know the process

    • one year ago
  4. zepdrix Group Title
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    You know the process, but you need assistance with the differentiation portion?

    • one year ago
  5. Dodo1 Group Title
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    Yes, when it comes to both chain rule and products. I get really confused

    • one year ago
  6. zepdrix Group Title
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    \[\large 2(x^2+y^2)^2=25(x^2−y^2)\]So we'll be differentiating this `implicitly`, meaning, `as is`. No solving for y ahead of time. When we execute the chain rule, a few y' will pop out here and there, and we'll want to solve for that variable. \(\large y'(-3,1)=m\). This is the m in our equation \(\large y=mx+b\). Hmm so let's see...

    • one year ago
  7. zepdrix Group Title
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    It appears we have no product rule which is nice.

    • one year ago
  8. Dodo1 Group Title
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    OK, Y7=(-3,1)=m. right?

    • one year ago
  9. zepdrix Group Title
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    We should be able to get a numerical value for m, no x's or y's in our m. I'm not really sure what that Y7 is.

    • one year ago
  10. Dodo1 Group Title
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    Ops my mistake

    • one year ago
  11. zepdrix Group Title
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    Taking the derivative of the left side gives us, `taking the derivative of the outermost function first, the square on the brackets`:\[\large 2\cdot2(x^2+y^2)^1\color{royalblue}{\frac{d}{dx}(x^2+y^2)}\]The blue is due to the chain rule. Understand that part?

    • one year ago
  12. Dodo1 Group Title
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    yes, so far good.

    • one year ago
  13. zepdrix Group Title
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    \[\large 2\cdot2(x^2+y^2)\color{royalblue}{(2x+2yy')}\]So we get something like this on the left side.

    • one year ago
  14. Dodo1 Group Title
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    i lied, I know product rule, power rule etc, but i dont know what the chain rule is

    • one year ago
  15. zepdrix Group Title
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    lol :)

    • one year ago
  16. zepdrix Group Title
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    mmmm i always have trouble explaining that one. It's very unintuitive.

    • one year ago
  17. zepdrix Group Title
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    hmm

    • one year ago
  18. Dodo1 Group Title
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    could you give me example of chain rule problem maybe?

    • one year ago
  19. zepdrix Group Title
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    The definition says this,\[\large \frac{d}{dx}f\left(g(x)\right) \qquad = \qquad f'(g(x))\cdot g'(x)\]When you take the derivative of this, a function within a function, you ~Differentiate the outer function, f in this case. ~Then you make a copy of the inner function, and multiply by it's derivative. Here is a silly example. \[\large \frac{d}{dx}\left(x^2\right)^{4}\] In this example we're thinking of \(\large g(x)=x^2\). And \(\large f(x)=x^4\). If you remember back to doing compositions of functions, you would understand that when we plug g(x) INTO f(x), our f(x), \(\large f(\color{royalblue}{x})=\left(\color{royalblue}{x}\right)^4\) will become, \(\large f(\color{royalblue}{g(x)})=\left(\color{royalblue}{g(x)}\right)^4\) which is \(\large f(\color{royalblue}{x^2})=\left(\color{royalblue}{x^2}\right)^4\). Applying the chain rule gives us,\[\large f'(g(x))=4\left(g(x)\right)^3 g'(x)\]See how the rule tells us to make a copy of the inner function and take it's derivative?\[\large f'(g(x))=4\left(x^2\right)^3 (x^2)'\]\[\large f'(g(x))=4\left(x^2\right)^3 (2x)\]

    • one year ago
  20. Dodo1 Group Title
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    Oo.. I see, Thank you. It was very clear! really cool :)

    • one year ago
  21. zepdrix Group Title
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    If you were to simplify this down, you would get,\[\large f'(g(x))=8x^7\] Which we know is correct, because we could have easily approached this example without using the chain rule.\[\large y=(x^2)^4=x^8\]And we know the derivative will match what we got from the chain rule example.

    • one year ago
  22. zepdrix Group Title
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    It's a weird concept :d you're like.. making copies of stuff.. but teachers never use that type of language to describe it. So I dunno lol

    • one year ago
  23. Dodo1 Group Title
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    no, you explain well. i always get confused and could not tell wether i should use chain or product.

    • one year ago
  24. zepdrix Group Title
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    So we applied the derivative of the left side,\[\large \frac{d}{dx}(thing)^2 \qquad \rightarrow \qquad 2(thing)(thing)'\] right? :O

    • one year ago
  25. Dodo1 Group Title
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    haha thats even more clear haha def, taking note

    • one year ago
  26. zepdrix Group Title
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    lol :d

    • one year ago
  27. Dodo1 Group Title
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    could we back to the first question?

    • one year ago
  28. Dodo1 Group Title
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    i took note

    • one year ago
  29. zepdrix Group Title
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    This one? :o \[\large 2(x^2+y^2)^2=25(x^2−y^2)\]

    • one year ago
  30. Dodo1 Group Title
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    yes, haha

    • one year ago
  31. Dodo1 Group Title
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    I see the chain rule now, (x^2+y^2)^2?

    • one year ago
  32. zepdrix Group Title
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    \[\large 2(x^2+y^2)\]That's the derivative of the outer function yes? The square on the brackets. But the chain rule is telling us to make a copy of the inside, and take it's derivative.\[\large 2(x^2+y^2)(x^2+y^2)'\]

    • one year ago
  33. zepdrix Group Title
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    \[\large 2(x^2+y^2)(2x+2y)\]So we should get somethingggggggg like this, yes?

    • one year ago
  34. Dodo1 Group Title
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    :O why two in the fron of the bracket? I understand the (2x+2y) tho

    • one year ago
  35. zepdrix Group Title
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    You applied the power rule to the outer function \(\large \left(\qquad \right)^2\)

    • one year ago
  36. Dodo1 Group Title
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    power rule and chain double? :O

    • one year ago
  37. zepdrix Group Title
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    Power rule the outside, make a copy of the inside, take derivative of copy.

    • one year ago
  38. Dodo1 Group Title
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    Ok, i understand

    • one year ago
  39. zepdrix Group Title
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    Since we took our derivative with respect to x, any time we differentiate a variable that IS NOT x, like y for example, we have to apply the chain rule, and multiply by y'.

    • one year ago
  40. zepdrix Group Title
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    \[\large 2(x^2+y^2)(2x+2y\color{royalblue}{y'})\]See this blue term? We took the derivative of a variable that is not x, namely y, so a y' popped out.

    • one year ago
  41. Dodo1 Group Title
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    mmmmmm.... Ok. I think :/

    • one year ago
  42. Dodo1 Group Title
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    so whenever its chain rule i put y' in the end?

    • one year ago
  43. zepdrix Group Title
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    No. Let's not think of that portion as chain rule, I think that's just going to confuse you. Whenever we take the derivative of y, we will multiply by y'. We don't throw it on the end, we always attach it to the y that we're taking the derivative of.

    • one year ago
  44. Dodo1 Group Title
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    right, Ok. i will remember that by my heart.

    • one year ago
  45. zepdrix Group Title
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    lol c:

    • one year ago
  46. Dodo1 Group Title
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    haha

    • one year ago
  47. Dodo1 Group Title
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    whats nxt stp?

    • one year ago
  48. zepdrix Group Title
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    Try taking the derivative of the right side, tell me what you get.

    • one year ago
  49. Dodo1 Group Title
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    Ok!

    • one year ago
  50. zepdrix Group Title
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    Ignore the 25 in front. No chain rule, just take the derivative of the stuff in the brackets.

    • one year ago
  51. Dodo1 Group Title
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    25(2x-2y)?

    • one year ago
  52. zepdrix Group Title
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    close, but you took the derivative of a y, you need to multiply that part by y'

    • one year ago
  53. Dodo1 Group Title
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    25(2x-2yy')?

    • one year ago
  54. Dodo1 Group Title
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    its still funny how i have to put y' in the end. is it for implicit derviation?

    • one year ago
  55. zepdrix Group Title
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    yes. If we had an equation like this,\[\large y=x^2\]The derivative gives us,\[\large y'=2x\]Ok that makes sense.. The y' is telling us what the derivative of y is. But what if we had something like this?\[\large y^2=x^2\]Taking the derivative,\[\large 2y=2x\] Hmm we don't have a derivative term do we?

    • one year ago
  56. Dodo1 Group Title
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    Ok, thats cool

    • one year ago
  57. Dodo1 Group Title
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    so 2(x^2+y~2)(2x+2y)=25(2x-2yy')

    • one year ago
  58. zepdrix Group Title
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    Wasn't there a 2 in front of the equation on the left at the start?? Or was that just a typo?

    • one year ago
  59. zepdrix Group Title
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    2(x^2+y^2)^2=25(x^2−y^2)

    • one year ago
  60. zepdrix Group Title
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    See that 2 at the front? That would give us an extra factor of 2 in our derivative.

    • one year ago
  61. Dodo1 Group Title
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    the question equation is 2(x^2+y^2)^2=25(x^2-y^2)

    • one year ago
  62. Dodo1 Group Title
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    so is it 2*2?

    • one year ago
  63. zepdrix Group Title
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    yes

    • one year ago
  64. Dodo1 Group Title
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    \[4(x^2+y^2)(2x+2y)=25(2x-2yy')\]

    • one year ago
  65. zepdrix Group Title
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    \[\large 4(x^2+y^2)(2x+2y\color{royalblue}{y'})=25(2x-2yy')\]Woops you missed a y'.

    • one year ago
  66. Dodo1 Group Title
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    ops! haha thats derivative of the first equation?

    • one year ago
  67. zepdrix Group Title
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    yes. We took the derivative of y on each side, so each y will get a y'

    • one year ago
  68. Dodo1 Group Title
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    Ok,

    • one year ago
  69. zepdrix Group Title
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    Hmmm from here, I think it's going to be easier to FIRST plug in your coordinate \(\large (-3,1\;)\) and THEN solve for y'.

    • one year ago
  70. Dodo1 Group Title
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    plug both number into the equation?

    • one year ago
  71. zepdrix Group Title
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    yes, -3 for all the x's. 1 in for all the y's.

    • one year ago
  72. zepdrix Group Title
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    and don't plug anything in for y'. That's what we'll be solving for.

    • one year ago
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